Question: Square $ABCD$ is inscribed in a circle with diameter $BD$ of length $2$ units. Segment $AB$ is the diameter of the half-circle on top of the square. What is the area of the shaded region, in square units? Express your answer as a common fraction. [asy] size(100); import graph; fill(Circle((.5,1),.5),mediumgray); fill(Circle((.5,.5),.5*sqrt(2)),white); draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); draw((0,0)--(1,1)); label("$D$",(0,0),SW); label("$C$",(1,0),SE); label("$B$",(1,1),NE); label("$A$",(0,1),NW); draw(Arc((.5,1),.5,0,180)); draw(Circle((.5,.5),.5*sqrt(2))); label("$2$",(.5,.5),NW); [/asy]
Answer: The shaded region is equal to area of the semicircle minus the area of the circular segment. $AB = \sqrt{2}$, so the area of the semicircle is $(1/2)(\sqrt{2}/2)^2\pi = \pi/4$. Let $O$ be the center of the large circle. The area of the circular segment is equal to the area of sector $ABO$ (which is a quarter-circle) minus the area of $\triangle ABO$. Thus the circular segment has area $(1/4)(1)^2\pi - (1/2)(1)(1) = \pi/4 - 1/2$. Hence the shaded region has area $\pi/4 - (\pi/4 - 1/2) = \boxed{1/2}$.